
\prob{0016}{两垂直求面积}

\begin{figure}[htbp]
  \centering
  \image{0016}
  \caption{0016：两垂直求面积} \label{fig:0016}
\end{figure}

如图~\ref{fig:0016}，在$\triangle ABC$中，$AD \perp BC$于$D$，$CE \perp AB$于$E$，$EM$平分$\angle BEC$交$AD$的延长线于$M$，连接$BM$、$CM$、$DM$，若$\angle CFD + \angle ABM = 180^\circ$，$2AE = 5BE$，$S_{\triangle AEF} = 5$，求$\triangle CEM$的面积。
\problabels{yellow/平面几何, green/面积问题}

\ans{$S_{\triangle CEM} = 25/3$}

\subsection{蝴蝶模型} \label{subsec:0016-butf}

基本思路：通过证明两个全等，得出$AC \parallel EM$以及$EF:CF$，然后运用蝴蝶模型求解。

\begin{align*}
  \because  {}& \angle CFD + \angle ABM = 180^\circ, \\
  & \angle CFD + \angle EFM = 180^\circ \\
  \therefore{}& \angle EBM = \angle EFM, \angle CEM = \frac12\angle BEC \\
  \because  {}& \text{$EM$ 平分 $\angle BEC$} \\
  \therefore{}& \angle BEM = \angle FEM \\
  \therefore{}& \triangle BEM \cong \triangle FEM\ \text{（证明省略）} \\
  \therefore{}& EB = EF \\
  \because  {}& 2AE = 5BE \\
  \therefore{}& 2AE = 5EF \\
  \because  {}& S_{\triangle AEF} = 5 \\
  \therefore{}& \frac12AE\cdot EF = 5 \\
  \therefore{}& AE = 5, EF = 2 \\
  \text{又}\because{}& CE \perp AB \\
  \therefore{}& \angle AEF = 90^\circ \\
  \therefore{}& \angle EAF + \angle AFE = 90^\circ \\
  \because  {}& \angle AEF + \angle CEB = 180^\circ \\
  \therefore{}& \angle CEB = 90^\circ \\
  \therefore{}& \angle AEF = \angle CEB, \angle CEM = 45^\circ \\
  \text{又}\because{}& AD \perp BC \\
  \therefore{}& \angle ADB = 90^\circ \\
  \therefore{}& \angle EAF + \angle CBE = 90^\circ \\
  \therefore{}& \angle AFE = \angle CBE \\
  \because  {}& \text{在 $\triangle AEF$ 与 $\triangle CEB$ 中} \\
  & \begin{cases}
    \angle AEF = \angle CEB \\
    EF = EB \\
    \angle AFE = \angle CBE
  \end{cases} \\
  \therefore{}& \triangle AEF \cong \triangle CEB \\
  \therefore{}& AE = CE \\
  \therefore{}& \angle ACE = 90^\circ - \frac12\angle AEC = 45^\circ \\
  \because  {}& \angle CEM = 45^\circ \\
  \therefore{}& \angle ACE = \angle CEM \\
  \therefore{}& AC \parallel EM \\
  \therefore{}& S_{\triangle AEM} = S_{\triangle CEM} \\
  \therefore{}& S_{\triangle AEF} = S_{\triangle CMF} = 5 \\
  \because  {}& AE = 5 \\
  \therefore{}& CE = 5 \\
  \because  {}& EF = 2 \\
  \therefore{}& CF = 3 \\
  \therefore{}& EF:CF = 2:3 \\
  \because  {}& S_{\triangle EMF}:S_{\triangle CMF} = EF:CF \\
  \therefore{}& S_{\triangle EMF}:5 = 2:3 \\
  \therefore{}& S_{\triangle EMF} = \frac{10}3 \\
  \because  {}& S_{\triangle CEM} = S_{\triangle EMF} + S_{\triangle CMF} \\
  \therefore{}& S_{\triangle CEM} = \frac{10}3 + 5 = \frac{25}3
\end{align*}

综上，$\triangle CEM$的面积为$\sfrac{25}3$。

\subsection{作垂线} \label{subsec:0016-vert}

\begin{figure}[htbp]
  \centering
  \image{0016-vert}
  \caption{\nameref{subsec:0016-vert}：通过作垂线构造等腰直角三角形，然后利用正弦定理和相似三角形求解。}
  \label{fig:0016-vert}
\end{figure}

基本思路：通过作垂线作出等腰直角三角形，然后根据相似三角形的比例关系结合等腰直角三角形的性质求解。

如图~\ref{fig:0016-vert}，作$MG \perp AB$于$G$。设$x = BG$。

\begin{align*}
  \because  {}& S_{\triangle BEM} = \frac12BE\cdot EM\sin\angle BEM, \\
  & S_{\triangle CEM} = \frac12CE\cdot EM\sin\angle CEM \\
  \therefore{}& S_{\triangle BEM}:S_{\triangle CEM} \\
  & = BE\sin\angle BEM:CE\sin\angle CEM \\
  \because  {}& EM\text{平分}\angle BEC \\
  \therefore{}& \angle BEM = \angle CEM \\
  \therefore{}& \sin\angle BEM = \sin\angle CEM \\
  \therefore{}& S_{\triangle BEM}:S_{\triangle CEM} = BE:CE \\
  \because  {}& \angle EAF = \angle GAM, \angle AEF = \angle G \\
  \therefore{}& \triangle EAF \sim \triangle GAM \\
  \therefore{}& AE:AG = EF:GM \\
  \because  {}& \triangle BEM \cong \triangle FEM \\
  & \text{（证法参见\nameref{subsec:0016-butf}）}, \\
  & S_{\triangle AEF} = 5 \\
  \therefore{}& AE = 5, EB = EF = 2 \\
  & \text{（证法参见\nameref{subsec:0016-butf}）} \\
  \because  {}& \angle GEM = \angle CEM, \angle FEG = 90^\circ \\
  \therefore{}& \angle GEM = 45^\circ \\
  \because  {}& \angle G = 90^\circ \\
  \therefore{}& \angle EMG = 45^\circ \\
  \therefore{}& \angle GEM = \angle EMG \\
  \therefore{}& GE = GM \\
  \because  {}& GE = x + 2 \\
  \therefore{}& GM = x + 2, AG = x + 7 \\
  \therefore{}& \frac5{x + 7} = \frac2{x + 2} \\
  \therefore{}& x = \frac{14}3 \\
  \therefore{}& GM = x + 2 = \frac{20}3 \\
  \therefore{}& S_{\triangle BEM} = \frac12EB\cdot GM = \frac{10}3 \\
  \because  {}& \triangle AEF \cong \triangle BEC \\
  & \text{（证法参见\nameref{subsec:0016-butf}）} \\
  \therefore{}& AE = CE = 5 \\
  \therefore{}& \frac{10}3:S_{\triangle CEM} = 2:5 \\
  \therefore{}& S_{\triangle CEM} = \frac{25}3 \\
\end{align*}

综上，$\triangle CEM$的面积为$25/3$。
